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Let $A, A^{-1} \in \mathbb{R}^{n \times n}$ be known matrices. Suppose we have an invertible matrix $B \in \mathbb{R}^{(n+1) \times (n+1)}$ of the following form:

$$B = \begin{bmatrix} A & b\\ b^T & 1 \end{bmatrix}$$

where $b$ is a column vector and $c$ is a row vector. How can I calculate matrix $B^{-1}$ from known matrices $A$ and $A^{-1}$? Can the Sherman–Morrison formula be applied here? If so, how?


As far as I understand, it can be applied if some perturbation is made to $A$. However, the problem here is that $B$ has a different shape than $A$. Appending $A$ with zero entries in the beginning will not work either because the same matrix with zeros added in the bottom row and the right column is not necessarily nonsingular.

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You know $\begin{pmatrix} A & 0 \\ 0 & 1 \end{pmatrix}^{-1} = \begin{pmatrix} A^{-1} & 0 \\ 0 & 1 \end{pmatrix}$, and from there you can make two successive rank-$1$ modifications, first adding $b$ along the last column, then $c$ along the last row. So, using Sherman–Morrison twice should work.

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    $\begingroup$ Presumably it is the same thing, but starting with $\pmatrix{A&b\\0&1\\}^{-1}=\pmatrix{A^{-1}&-A^{-1}b\\0&1\\}$ is good. $\endgroup$ – Brendan McKay Jun 7 '20 at 4:37
  • $\begingroup$ I actually have found a ready to use formula derived from Sherman-Morrison. I put the answer here: github.com/Bigpig4396/… $\endgroup$ – Manh Khôi Duong Jun 8 '20 at 3:34

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